If we have elements in the lie algebra $a, b \in \mathfrak{g}$, when is it true that $\exp(a) \exp(b) = \exp(b) \exp(a)$?
Naively, I believe that if $a$ and $b$ are linearly independent, then their exponentials can commute. I'm not sure how to prove this, however.
In the case of matrix groups, I believe the correct condition is that when $a$ and $b$ commute, then we have that $\exp(a) \exp(b) = \exp(b) \exp(a)$, since the matrices $a, b$ will behave just like real numbers. But this "feels wrong", since I am invoking the multiplicative structure of $a$ and $b$. But as elements of the lie algebra, all I have is the vector space structure!
I suppose the way to correct this is to use the lie bracket structure, and state that if $[a, b] = 0$, then the exponentials commute? I am still looking for a proof, though. I know a "nuke proof" that invokes BCH. If $[X, Y] = 0$, then we can solve for $\exp(a) \exp(b) = exp(z)$, where:
$$ z = a + b + \frac{1}{2}[a, b] + \dots \text{expressions involving }[X, Y] \\ z = a + b + 0 = a + b $$
So this gives me $\exp(a) \exp(b) = \exp(a + b) = \exp(b + a) = \exp(b) \exp(a)$.
But this is way too much firepower to prove something that feels elementary; BCH is complicated. What is a simple proof of the above fact?